Respuesta :
Explanation:
The given data is as follows.
Initial temperature ([tex]T_{1}[/tex]) = 400 K
Final temperature ([tex]T_{2}[/tex]) = 298 K
Process [tex]P \nu^{1.2}[/tex] = C
For [tex]CO_{2}[/tex], according to the properties of various substances the values are as follows.
R = 0.1889 kJ/kg K
[tex]C_{p}[/tex] = 0.846 kJ/kg K
Therefore, calculate the final pressure as follows.
[tex]\frac{P_{2}}{P_{1}} = (\frac{T_{2}}{T_{1}})^{\frac{n}{n-1}}[/tex]
[tex]\frac{P_{2}}{6} = (\frac{298}{400})^{\frac{1.2}{1.2-1}}[/tex]
[tex]P_{2}[/tex] = 1.025 bar
Hence, final pressure is 1.025 bar.
Now, work done will be calculated as follows.
W = [tex]\frac{P_{1}\nu_{1} - P_{2}\nu_{2}}{n - 1}[/tex]
= [tex]\frac{m(RT_{1} - RT_{2})}{n - 1}[/tex]
= [tex]\frac{1 \times 0.1889(400 - 298)}{(1.2 - 1)}[/tex]
= 96.339 kJ/kg
Hence, work done is equal to 96.339 kJ/kg.
Formula to calculate the change in internal energy is as follows.
[tex]\Delta U = C_{v}(T_{2} - T_{1})[/tex]
= 0.657 (298 - 400)
= -67.014 kJ/kg
Now, transfer of heat will be calculated as follows.
Q = [tex]\Delta U + W[/tex]
= -67.014 kJ/kg + 96.339 kJ/kg
= 29.325 kJ/kg
Therefore, amount of heat transfer is 29.325 kJ/kg.
